Undergrad What kind of tensor is the gradient of a vector Field?

[Orodruin]

(ie a metric tensor field)

No, not necessarily a metric tensor field. A non-degenerate bilinear form. This is not the same thing. A metric tensor is a bilinear form but a bilinear form is not necessarily a metric (or even a pseudo-metric). Again, the typical example would be the (fully anti-symmetric) symplectic form in symplectic geometry.

Please pay attention to the bold words

Please pay attention to the bold words:

All tangent spaces to all points on a manifold are isomorphic

 

[lavinia]

A vector space and its dual differ in the way their elements transform under linear maps. If a linear map $L$ maps the vector space $V$ into the vector space $W$ then by definition elements of $V$ are sent to elements of $W$.

However elements of the dual space $V^{*}$ are not sent to elements of $W^{*}$. Instead it is the other way around. Elements of $W^{*}$ are sent to elements of $V^{*}$.

Schematically this transformation difference can be represented by two diagrams:

$L:V→V$ $L^{*}:W^{*}→V^{*}$

This difference in transformation direction distinguishes a vector space and its dual. I would guess that this automatically excludes the possibility of a natural transformation between the identity functor and the "dual"" functor that maps each vector space to its dual and each linear map to its induced map on dual spaces. A natural transformation would seem always to preserve the direction of transformation. @fresh_42 is this correct?

This is a purely algebraic consideration and does not require manifolds or tensors or anything else. However, on a smooth manifold, the differential of a smooth map $f:M→N$ between two manifolds $df:TM→TN$ maps tangent vector in $TM$ to tangent vectors in $TN$ while the dual of the differential $df^{*}:TN^{*}→TM^{*}$ maps $TN^{*}$ back in the opposite direction to $TM^{*}$. The direction of $df$ is called "covariant" while the direction of $df^{*}$ is called "contravariant".

In the language of what is sometimes called "abstract nonsense" the differential is a covariant functor from the category of smooth manifolds and smooth maps to the category of smooth vector bundles and smooth vector bundle morphisms while the dual of the differential defines a contravariant functor to the category of smooth vector bundles and smooth vector bundle morphisms. This is the origin of the terms "push forward" and "pullback".

Notes:
-The coordinates of a vector with respect to a basis are dual vectors when they are viewed as linear functionals that assign a coordinate to each vector. Similarly, the coordinates of dual vectors are dual dual vectors and under the canonical isomorphism are identified with vectors.

 

[fresh_42]

A natural transformation would seem always to preserve the direction of transformation. @fresh_42 is this correct?

If I read nLab correctly, and there is no better online source I am aware of when it comes to homological algebra then there is no distinction between covariant or contravariant functors in the definition of a natural transformation between categories, i.e. it is valid in both cases.

Assume we would have two different functors $\tau,\dagger \, : \,\text{Vec}\longrightarrow \text{Vec}^*.$ A transformation $\alpha \, : \, \tau \longmapsto \dagger$ is natural if it is compatible with the morphisms: Say $\varphi : V^\tau \longrightarrow W^\tau$ and $\psi : V^\dagger \longrightarrow W^\dagger$ then we demand that
$$
\alpha (W)\circ \varphi^\tau = (V^\tau\stackrel{\varphi^\tau}{\longrightarrow }W^\tau \stackrel{\alpha }{\longrightarrow } W^\dagger) =(V^\tau\stackrel{\alpha }{\longrightarrow }V^\dagger \stackrel{\psi^\dagger}{\longrightarrow }W^\dagger)= \psi^\dagger \circ \alpha (V)
$$
Hence being natural says something about the mapping $\tau \longmapsto \dagger$ rather than the direction of $\varphi$ or $\psi.$ I think that makes sense as we are interested how $V^*\cong V^\tau$ and $V^*\cong V^\dagger$ are related, and not how morphisms in either category are defined. If I'm not mistaken, then $\varphi,\psi \, : \,W\longrightarrow V$ is not ruled out.

There is another interesting note on that page:
A transformation which is natural only relative to isomorphisms may be called a canonical transformation.

At least this explains the difficulty to distinguish the terms with the example $\text{Vec}$ and $\text{Vec}^*.$

 

[lavinia]

If I read nLab correctly, and there is no better online source I am aware of when it comes to homological algebra then there is no distinction between covariant or contravariant functors in the definition of a natural transformation between categories, i.e. it is valid in both cases.

Assume we would have two different functors $\tau,\dagger \, : \,\text{Vec}\longrightarrow \text{Vec}^*.$ A transformation $\alpha \, : \, \tau \longmapsto \dagger$ is natural if it is compatible with the morphisms: Say $\varphi : V^\tau \longrightarrow W^\tau$ and $\psi : V^\dagger \longrightarrow W^\dagger$ then we demand that
$$
\alpha (W)\circ \varphi^\tau = (V^\tau\stackrel{\varphi^\tau}{\longrightarrow }W^\tau \stackrel{\alpha }{\longrightarrow } W^\dagger) =(V^\tau\stackrel{\alpha }{\longrightarrow }V^\dagger \stackrel{\psi^\dagger}{\longrightarrow }W^\dagger)= \psi^\dagger \circ \alpha (V)
$$

Right. What I meant to say more clearly was there is no idea of a natural transformation between a covariant functor and a contravariant functor.

The thinking was whether one could choose isomorphisms $J_{V}$ between each vector space $V$ an its dual so that for every linear map $L:V→W$, $L^{*} \circ J_{W} \circ L =J_{V}$. This cannot be done by choosing a basis for each vector space.

This seems to illustrate a difference between the idea of a natural/canonical isomorphism and a natural transformation.

 

[S.G. Janssens]

Why not?

As an example, take a Banach space that is not reflexive but still linearly isomorphic to its second continuous dual space.

 

[martinbn]

As an example, take a Banach space that is not reflexive but still linearly isomorphic to its second continuous dual space.

How is that in contradiction to what I said!

 

[S.G. Janssens]

How is that in contradiction to what I said!

Is the above a question?

If yes, then it can be answered by looking up reflexivity and, for example, James' theorem.